The mechanical energy of the girl will be conserved because the system is isolated and the initial potential energy will be equal to final kinetic energy.
<h3>
What is the law of conservation of energy?</h3>
The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.
The change in the potential energy of the launched from a height into the pool without friction from the given height h is calculated by applying the following kinematic equation.
ΔP.E = ΔK.E
where;
- ΔP.E is change in potential energy of the child
- ΔK.E is change in the kinetic energy of the child
mghf - mghi = ¹/₂mv² - ¹/₂mu²
where;
- m is the mass of the girl
- g is acceleration due to gravity
- hi is the initial height of the girl
- hf is the final height when she is launched into the pool
- u is the initial velocity
- v is the final velocity of the girl
Thus, for every closed or isolated system such as this case, mechanical energy is always conserved because the initial potential energy of the girl will be converted into her final kinetic energy.
Learn more about conservation of mechanical energy here: brainly.com/question/332163
#SPJ1
I attached a picture of the diagram associated with this question.
Now,
When we check the vertical components of the tension in the rope, we will find that we have two equal components acting upwards.
These two components support the weight and each of them has a value of TcosΘ
The net force acting on the body is zero.
Fnet=Force of tension acting upwards-Force due to weight acting downwards
0 = 2TcosΘ -W
W = 2TcosΘ
T = W / 2cosΘ
Answer:
The gravitational acceleration of the planet is, g = 8 m/s²
Explanation:
Given data,
The distance the object falls, s = 144 m
The time taken by the object is, t = 6 s
Using the III equations of motion
S = ut + ½ gt²
∴ g = 2S/t²
Substituting the given values,
g = 2 x 144 /6²
= 8 m/s²
Hence, the gravitational acceleration of the planet is, g = 8 m/s²
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where
is the density of freshwater which has a constant value

Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as


So 
=> 
Now the work done is mathematically represented as

![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
